1. The document discusses the gyroscopic effect, which is the tendency of a spinning object to resist any change to its axis of rotation. It explains how a gyroscopic couple is generated when a spinning object experiences precession. 2. Key applications of gyroscopic effect discussed include aeroplanes, ships, vehicles. For aeroplanes, the effect of the spinning engine/propeller is to change the plane's attitude during turns. For ships steering or pitching, it causes the bow/stern to raise or lower. There is no effect during ship rolling. 3. Sample calculations are provided to determine the gyroscopic couple generated for different rotating objects like engines, flywheels, and to analyze their effect

1. ME 010 601 Mechanics of Machines
Module - 4
Gyroscope
Binil Babu
Department of Mechanical Engineering
SNMIMT, Maliankara.

2. Contents
1 Overview 1
2 Principle 1
3 Precessional Angular Motion 2
3.1 Gyroscopic couple . . . . . . . . . . . . . . . . . . . . . . . . . 3
4 Effect of gyroscopic couple on aeroplanes 4
5 Effect of gyroscopic couple on ships 8
5.1 Effect of gyroscopic couple on a naval ship during steering . . 8
5.2 Effect of gyroscopic couple on a naval ship during pitching . . 10
5.3 Effect of Gyroscopic Couple on a Naval Ship during Rolling . . 10
6 Effect of gyroscopic couple on four wheel 13
7 Stability of a Two Wheel Vehicle Taking a Turn 16
8 Stability of a two wheel vehicle taking a turn 19
9 Effect of gyroscopic couple on a disc rigidly fixed at a certain
angle to a rotating shaft 21
10 Previous year university questions 24
11 Reference 25
i

3. GYROSCOPE
1 Overview
The earliest work and studies on gyroscopes were carried out by Newton.
The mathematical foundation of principle of foundation of principles of gyro-
scopic motion done by Euler.He derived a set of dynamical equations relating
applied moment,angular acceleration and angular velocity. In many cases the
rotating body/element forced to turn about there axes other than their own
axes of rotation, and gyroscopic effects are set up.
2 Principle
When a body moves along a curved path with a uniform linear velocity, a
force in the direction of centripetal acceleration (known as centripetal force)
has to be applied externally over the body, so that it moves along the required
curved path. This external force applied is known as active force.
When a body, itself, is moving with uniform linear velocity along a
circular path, it is subjected to the centrifugal force(magnitude is equal to
centripetal force) radially outwards. This centrifugal force is called reactive
force. The action of the reactive or centrifugal force is to tilt or move the
body along radially outward direction.Whenever the effect of any force or
couple over a moving or rotating body is to be considered, it should be with
respect to the reactive force or couple and not with respect to active force
or couple. Simply the effect on a moving body due to this reactive
force/couple is called gyroscopic effect /couple.
1

4. 3 Precessional Angular Motion
Figure 1: Precessional Angular Motion
Consider a disc, in the above figure spinning about the axis OX (known
as axis of spin) in anticlockwise when seen from the front, with an angular
velocity ω in a plane at right angles to the paper.
After a short interval of time δt let the disc be spinning about the new axis
of spin OX at an angle δθ and angular velocity (ω + δω) initial angular
velocity of the disc (ω) is represented by vector ox; and the final angular
velocity of the disc (ω + δω) is represented by vector ox as shown in Fig (b).
The vector xx represents the change of angular velocity in time δt i.e. the
angular acceleration of the disc. This may be resolved into two components,
one parallel to ox and the other perpendicular to ox.
Component of angular acceleration in the direction of ox,
αt = xr
δt
= or−ox
δt
= ox cos(δθ)−ox
δt
After solving the equation we will get αt = δω
δt
, when δt → 0,
αt = dω
dt
Component of angular acceleration in the direction perpendicular to ox,
αc = rx
δt
= (ω+δω)sinδθ
δt
αc = ωδθ
δt
Lt δt → 0, αc = ωωp
2

5. Total angular acceleration of the disc = αt + αc
ωp is called angular velocity of precession.The axis of precession is perpen-
dicular to the plane in which the axis of spin is going to rotate. The angular
motion of the axis of spin about the axis of precession is known as precessional
angular motion.
3.1 Gyroscopic couple
Figure 2: Gyroscopic couple
Consider a disc spinning with an angular velocity ω rad/s about the axis
of spin OX, in anticlockwise direction when seen from the front, as shown in
Fig.(a). Since the plane in which the disc is rotating is parallel to the plane
YOZ, therefore it is called plane of spinning. The plane XOZ is a horizontal
plane and the axis of spin rotates in a plane parallel to the horizontal plane
about an axis OY. In other words, the axis of spin is said to be rotating or
processing about an axis OY. In other words, the axis of spin is said to be
rotating or processing about an axis OY (which is perpendicular to both the
axes OX and OZ) at an angular velocity ωp rad/s. This horizontal plane
XOZ is called plane of precession and OY is the axis of precession.
3

6. I = Mass moment of inertia of the disc about OX, and
ω = Angular velocity of the disc.
Therefore Angular momentum of the disc = I.ω
The axis of spin OX is also rotating anticlockwise when seen from the
top about the axis OY. Let the axis OX is turned in the plane XOZ through
a small angle δθ radians to the position OX’ , in time δt seconds. Assuming
the angular velocity ω to be constant, the angular momentum will now be
represented by vector ox .
Change in angular momentum ox –ox = xx = oxδθ,
and the rate of change of angular momentum = I.ωδθ
δt
Since the rate of change of angular momentum will result by the applica-
tion of a couple to the disc, therefore the couple applied to the disc causing
precession.
C = Lt δt → 0 I.ω.δθ
δt
= I.ω.ωp
C = I.ω.ωp
4 Effect of gyroscopic couple on aeroplanes
Let engine or propeller rotates in the clockwise direction when
seen from the rear or tail end and the aeroplane takes a turn to
the left.
ω = Angular velocity of the engine in rad/s,
m = Mass of the engine and the propeller in kg,
k = Its radius of gyration in metres,
4

7. I = Mass moment of inertia of the engine and the propeller
in kg-m2 = m.k2
,
v = Linear velocity of the aeroplane in m/s,
R = Radius of curvature in metres, and
ωP = Angular velocity of precession , v
R rad/s
Gyroscopic couple acting on the aeroplane , C = I.ω.ωp
Figure 3: Effect of gyroscopic couple on aeroplanes
Figure 4: Effect of gyroscopic couple on aeroplanes
Before taking the left turn, the angular momentum vector is
represented by ox. When it takes left turn, the active gyroscopic
5

8. couple will change the direction of the angular momentum vec-
tor from ox to ox as shown in figure. The vector xx in the
limit represents the change of angular momentum or the active
gyroscopic couple and is perpendicular to ox.
The reactive gyroscopic couple (equal in magnitude of active
gyroscopic couple) will act in the opposite direction (i.e. in the
anticlockwise direction) and the effect of this couple is, therefore,
to raise the nose and dip the tail of the aeroplane.When
the aeroplane takes a right turn under similar conditions as dis-
cussed above, the effect of the reactive gyroscopic couple will be
to dip the nose and raise the tail of the aeroplane.
• When the engine or propeller rotates in anticlockwise di-
rection when viewed from the rear or tail end and the aero-
plane takes a left turn, then the effect of reactive gyroscopic
couple will be to dip the nose and raise the tail of the
aeroplane.
• When the aeroplane takes a right turn ,When the engine
or propeller rotates in anticlockwise direction when viewed
from the rear or tail end the effect of reactive gyroscopic
couple will be to raise the nose and dip the tail of the
aeroplane.
• When the engine or propeller rotates in clockwise direction
when viewed from the front and the aeroplane takes a left
turn, then the effect of reactive gyroscopic couple will be
to raise the tail and dip the nose of the aeroplane.
• When the aeroplane takes a right turn When the engine or
propeller rotates in clockwise direction when viewed from
the front and the aeroplane takes a right turn, the effect of
reactive gyroscopic couple will be to raise the nose and
dip the tail of the aeroplane.
6

9. • The gyroscopic effect for propeller rotates in clockwise di-
rection when looking from rear end and takes left turn will
be same for propeller rotates in anticlockwise direction look-
ing from the rear end and take right turn.
Problem.1
An aeroplane makes a complete half circle of 50 metres ra-
dius, towards left, when flying at 200 km per hr. The rotary
engine and the propeller of the plane has a mass of 400 kg and
a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m.
clockwise when viewed from the rear. Find the gyroscopic cou-
ple on the aircraft and state its effect on it.
Given data: R = 50 m, v = 200 km/hr = 55.6 m/s, m =
400 kg, k = 0.3 m, N = 2400 rpm.
Gyroscopic couple, C = IωωP
I, Mass moment of inertia = mk2
= 400×0.32
= 36 kg − m2
ω, Angular velocity = 2πN
60 = 2π×2400
60 = 251 rad/sec
ωP , Precessional angular velocity = v
R = 55.6
50 = 1.11 rad/sec
C = IωωP = 36 × 251 × 1.11 = 10.046 kN-m.
The effect of reactive gyroscopic couple is ”Raise the nose
Dip the tail”
7

10. 5 Effect of gyroscopic couple on ships
The top and front views of a naval ship are shown in Figure
below. The fore end of the ship is called bow and the rear
end is known as stern or aft. The left hand and right hand
sides of the ship, when viewed from the stern are called port
and star-board respectively. We shall now discuss the effect
of gyroscopic couple on the naval ship in the following three
cases: 1. Steering, 2. Pitching, and 3. Rolling
Figure 5: Terms used in naval ship
5.1 Effect of gyroscopic couple on a naval ship during
steering
Steering is the turning of a complete ship in a curve towards
left or right, while it moves forward. Consider the ship taking
8

11. Figure 6: Effect of gyroscopic couple on a naval ship during steering
Figure 7: Steering of ship
a left turn, and rotor rotates in the clockwise direction when
viewed from the stern, as shown in figure below. The effect of
gyroscopic couple on a naval ship during steering taking left or
right turn may be obtained in the similar way as for an aeroplane
as discussed in previous case.
When the rotor of the ship rotates in the clockwise direction
when viewed from the stern, it will have its angular momentum
vector in the direction ox as shown in figure. As the ship steers
to the left, the active gyroscopic couple will change the angular
momentum vector from ox to ox . The vector xx now represents
the active gyroscopic couple and is perpendicular to ox. Thus
the plane of active gyroscopic couple is perpendicular to xx
and its direction in the axis OZ for left hand turn is clockwise
as shown in above figure. The reactive gyroscopic couple of
the same magnitude will act in the opposite direction (i.e. in
anticlockwise direction). The effect of this reactive gyroscopic
9

12. couple is to raise the bow and lower the stern.When the
ship steers to the right under similar conditions as discussed
above, the effect of the reactive gyroscopic couple, as shown in
Fig.7, will be to raise the stern and lower the bow.
5.2 Effect of gyroscopic couple on a naval ship during
pitching
Pitching is the movement of a complete ship up and down in a
vertical plane about transverse axis, as shown in figure 7. In this
case, the transverse axis is the axis of precession. The pitching of
the ship is assumed to take place with simple harmonic motion
i.e. the motion of the axis of spin about transverse axis is simple
harmonic.
Figure 8: Effect of gyroscopic couple on a naval ship during pitching
5.3 Effect of Gyroscopic Couple on a Naval Ship dur-
ing Rolling
The effect of gyroscopic couple to occur, the axis of precession
should always be perpendicular to the axis of spin. If, however,
the axis of precession becomes parallel to the axis of spin, there
10

13. will be no effect of the gyroscopic couple acting on the body
of the ship. In case of rolling of a ship, the axis of precession
(i.e. longitudinal axis) is always parallel to the axis of spin for
all positions. Hence, there is no effect of the gyroscopic couple
acting on the body of a ship.
Problem 2. A turbine rotates at 1200 rpm CW when look-
ing from stern. The sea vessel pitches at 1.2 rad/s. Find out
the gyroscopic couple transmitted to the hull when bow rises.
Given data
N = 1200 rpm
ω = 2πN
60 = 125.66 rad/s
ωp = 1.2 rad/s
m = 950 kg
k = 300 mm = 0.3 m
Sol:
I = mk2
= 950 × 0.32
= 85.5 kg − m2
C = Iωωp = 85.5 × 125.66 × 1.2 = 12.892kN − m
The effect of gyroscopic couple is tends to move the
ship towards starboard .
Problem 3. A flywheel with mass 20 kg,radius of gyration
300 mm spinning at 500 rpm about a horizontal axis. The fly
wheel is suspended at a point 250 mm from the plane of rotation
of flywheel. Find the rate of precession of the fly wheel.
11

14. Given data:
m = 20 kg, k = 300 mm = 0.3 m, N = 500 rpm, l = 250 mm
= 0.25 m
Soln:
ω = 2πN
60 = 2π×500
60 = 52.35rad/s
I = mk2
= 20 × 0.32
= 1.8kg − m2
C = mgl = 20×9.81×0.25 = 49 N-m, we know that,
C = Iωωp ie; 49 N-m = 1.8 × 52.35 × ωp
ωp = 49
1.8×52.35 = 0.52 rad/s
12

15. 6 Effect of gyroscopic couple on four wheel
Figure 9: Effect of gyroscopic couple on four wheel
m = mass of vehicle in kg
W = weight of vehicle in N
v = Linear velocity in m/s
R = radius of curvature in m
ωp = precession velocity in rad/s
ωW = Angular velocity of wheel in rad/s
ωW = Angular velocity of rotating parts of engine in rad/s
IW = Mass moment of inertia of wheels
IE = Mass moment of inertia of rotating parts of engine
13

16. G = Gear ratio = ωE
ωW
In this case we are assuming that the weight of vehicle is
equally distributed through 4 wheels. Therefore weight dis-
tributed on each wheel is given by W
4 . The reaction between
road and wheels due this load/weight acting in upward direc-
tion on all wheels.
Effect due to gyroscopic couple.
Since the vehicle takes a turn towards left due to the preces-
sion and other rotating parts, therefore a gyroscopic couple will
act. The velocity of precession,is ωp = v
R.Therefore gyroscopic
couple due to 4 wheels,
CW = 4IW ωW ωp
The gyroscopic couple due to the rotating parts of the en-
gine, is given by equation
CE = IEωEωp
The net gyroscopic couple is given by C = CW ± CE,
C = 4IW ωW ωp ± IEGωW ωp
C = ωW ωp(4IW ± IEG) Positive sign used when both wheel
and rotating parts of engine rotates in same direction and neg-
ative sign used when wheel and rotating parts of engine rotates
in opposite direction.
Due to the gyroscopic couple, vertical reaction on the road sur-
14

17. face will be produced. The reaction will be vertically upwards on
the outer wheels and vertically downwards on the inner wheels.
Let the magnitude of this reaction at the two outer or inner
wheels be P Newtons. Then P × x = CorP = C
x ,therefore ver-
tical reaction at each of the outer or inner wheels, P
2 = C
2x
Effect of centrifugal couple. When CE > CW, then C will be
–ve. Thus the reaction will be vertically downwards on the outer
wheels and vertically upwards on the inner wheels
Since the vehicle moves along a curved path, therefore centrifu-
gal force will act outwardly at the centre of gravity of the vehicle.
The effect of this centrifugal force is also to overturn the vehicle.
We know that centrifugal force,
FC = mv2
R
The couple tending to overturn the vehicle or overturning
couple is,
CO = FC × h = mv2
R × h. This overturning couple is balanced
by vertical reactions, which are vertically upwards on the outer
wheels and vertically downwards on the inner wheels. Let the
magnitude of this reaction at the two outer or inner wheels be
Q. Then
Q × x = CO , Q = CO
x
Q = mv2
h
2Rx
therefore total vertical reaction on each outer wheel is given
by, PO = W
4 + P
2 + Q
2
15

18. And reaction on inner wheel PI = W
4 − P
2 − Q
2
To have contact between inner wheels and road,the following
condition must be satisfy. ie; sum of P
2 and Q
2 should be less
than W
4
7 Stability of a Two Wheel Vehicle Taking a
Turn
Figure 10: Stability of a Two Wheel Vehicle Taking a Turn
m = Mass of the vehicle and its rider in kg,
W = Weight of the vehicle and its rider in newtons = m.g,
h = Height of the centre of gravity of the vehicle and rider,
rW = Radius of the wheels,
R = Radius of track or curvature,
16

19. IW = Mass moment of inertia of each wheel,
IE = Mass moment of inertia of the rotating parts of the
engine,
ωW = Angular velocity of the wheels,
ωE = Angular velocity of the engine,
G = Gear ratio = ωE / ωW ,
v = Linear velocity of the vehicle = ωW × rW ,
θ = Angle of heel. It is inclination of the vehicle to the ver-
tical for equilibrium
Effect of gyroscopic couple,
we know that v = ωW × rW , ωW = v
rW
ωE = G.ωW = G. v
rW
Total (I × ω) = 2IW × ωW ± IE × ωE
= 2IW × v
rW
± IE × G × v
rW
= r
ωW
(2IW ± G.IE)
ωp = v
R
The fig.show that when the wheels move over the curved path,
the vehicle is always inclined at an angle θ with the vertical
plane as shown in Fig.This angle is known as angle of heel. In
other words, the axis of spin is inclined to the horizontal at an
17

20. angle θ. Thus the angular momentum vector Iω due to spin
is represented by OA inclined to OX at an angle θ. But the
precession axis is vertical. Therefore the spin vector is resolved
along OX.Therefore the gyroscopic couple is,
C1 = Iωcosθ × ωp = v2
R.rW
(IW ± G.IE)cosθ
The gyroscopic couple will act over the vehicle outwards i.e.
in the anticlockwise direction when seen from the front of the
vehicle. The tendency of this couple is to overturn the vehicle
in outward direction.
Effect of centrifugal couple
When vehicle moves along a curved path centrifugal force will
act outward direction at centre of gravity of the vehicle. This
centrifugal force tries to overturn the vehicle.
Centrifugal force FC = m×2
v
R
The overturning couple ,CO = FC × h = m.v2
R × h
This overturning couple is balanced by vertical reactions, which
are vertically upward on outer wheels and vertically downward
ion inner wheels. let the magnitude of this reaction force on
”two” inner or outer wheels be ”Q”
Qtimesx = CO or Q = CO
x = m.v2
.h
R.x
Vertical reaction at each inner or outer wheel,
Q
2 = m.v2
.h
2.R.x
18

21. Total vertical reaction at each of outer wheel,
PO = W
4 + P
2 + Q
2
Total vertical reaction on inner wheel,
PI = W
4 − P
2 − Q
2
When the value of PI is negative or zero the wheel will lift from
the road ie; the vehicle overturn while moving through a curved
path. In order to have the contact between the inner wheels and
the ground, the sum of P
2 and Q
2 must be less than W
4 .
8 Stability of a two wheel vehicle taking a
turn
Figure 11: Stability of a two wheel vehicle taking a turn
m = Mass of the vehicle and its rider in kg,
W = Weight of the vehicle and its rider in newtons = m.g,
19

22. h = Height of the centre of gravity of the vehicle and rider,
rW = Radius of the wheels
R = Radius of track or curvature,
IW = Mass moment of inertia of each wheel
IE = Mass moment of inertia of the rotating parts of the engine,
ωW = Angular velocity of the wheels,ωW = v
rW
ωE = Angular velocity of the engine
G = Gear ratio = ωE
ωW
v = Linear velocity of the vehicle = ωW × rW
θ = Angle of heel. Inclination of vehicle to vertical axis for
equilibrium.
Effect of gyroscopic couple.
C = Iωωp, Total (I×ωp) = 2IW × ωW ± IE × ωE,
C = 2IW × ωW ± IE × G.ωW
= 2IW × v
rW
± IE × G × v
rW
= v
rW
(2IW ± G.IE)
Precessional angular velocity ωp = v
R
Gyroscopic couple
20

23. The axis of spin is inclined to horizontal plane at an angle
θ. Since the precession axis is vertical we need to take the hori-
zontal component of spin angular velocity ie; ω cos θ. Therefore
gyroscopic couple
C1 = I.ω cos θ × ωp
Effect of centrifugal couple.
FC = m.v2
R acts along cog in outward direction;
centrifugal couple C2 = FC × h. cos θ, this couple have a ten-
dency to overturn the vehicle.Therefore total overturning couple,
CO = Gyroscopic couple + Centrifugal couple
= v2
R.rW
(2IW + G.IE) cos θ + m.v2
R × h. cos θ
= v2
R [2IW +G.IE
rW
+ m.h] cos θ
Balancing couple = m.g.hsin θ. It acts in clockwise direction
when seen from the front of the vehicle. For stability the over-
turning couple should be equal to balancing couple.
9 Effect of gyroscopic couple on a disc rigidly
fixed at a certain angle to a rotating shaft
Consider a disc rigidly fixed on shaft such that polar axis of disc
makes an angle θ with shaft axis. The shaft rotates at angular
velocity ω in CW direction when looking from front.
21

24. Figure 12: Effect of gyroscopic couple on a disc rigidly fixed at a certain
angle to a rotating shaft
OX is the axis of shaft
OP is the polar axis of disc
OD horizontal diametral axis
Let m,r,l be the mass ,radius, and width/thickness of disc.
Mass moment of inertia of disc about polar axis IP = mr2
2
Mass moment of inertia about diametral axis ID = m( l2
12+r2
4 ),
the value of l can be neglected when the disc is thin, then
ID = mr2
4
Angular velocity of disc about polar axis = ω cos θ
Angular velocity about diametral axis = ω sin θ ( Here di-
ametral axis is the precession axis,so the angular velocity about
diametral axis is precessional velocity)
Now the gyroscopic couple acting on disc = IP ω cos θ.ω sin θ
22

25. = 1/2 × IP .ω2
sin 2θ (2 sin θ cos θ = sin 2θ)
This gyroscopic couple is to turn the disc in anticlockwise
when viewed from top about an axis through ”O”.
Now consider OD,the diametral axis as the axis of spin and
polar axis OP as the axis of precession,
The angular velocity of spin = ω. sin θ
Precessional velocity,ie; the velocity about polar axis = ω. cos θ
If ID is the mass moment of inertia of the disc about diame-
tral axis OD, then gyroscopic couple acting on disc,
CD = ID.ω sin θ.ω cos θ
= 1
2 × ID.ω2
sin 2θ
The effect of this couple is opposite to CP , therefore resul-
tant couple acting on disc,
C = CP − CD = 1
2 × ω2
sin 2θ.(IP − ID)
After substituting the value of IP and ID,
C = m
8 × ω2
.r2
. sin 2θ
This resultant gyroscopic couple will act in the anticlockwise
direction as seen from the top.
Problem 1. A shaft carries a uniform thin disc of 0.6 m di-
ameter and mass 30 kg. the disc of is out of truth and makes
23

26. an angle of 1o
with a plane at right angles to the axis of the
shaft. find the gyroscopic couple acting on the bearing when
shaft rotates at 1200 rpm.
Given data :-
d = 0.6 m, r = 0.3 m
m = 30 kg
θ = 1o
N = 1200 rpm, ω = 2πN
60 = 2π×1200
60 = 125.7rad/s
Gyroscopic couple acting on bearings, C = m
8 × ω2
r2
sin 2θ
= 30
8 × (125.7)2
(0.3)2
sin 2o
= 186 N-m
10 Previous year university questions
1.What is the magnitude of gyroscopic couple when ship rolls
about its roller axis.
2.What is mean by gyroscopic couple.?Derive a relation for its
magnitude.
3.List all devices/machines working on the principle of gyro-
scopic couple.
4.With neat sketch,explain working of a shaft -disc system.
5.Explain pitching,rolling, and yawing of ships with neat dia-
gram.
6.Explain gyroscopic effect on a two wheeled vehicle.
7.Discuss gyroscopic effect on sea vessels.
24

27. 11 Reference
1.https://www.youtube.com/watch?v=ekzwbt3hu2k
2.Theory of Machines - S S Rattan
3.Theory of Machines - Khurmi& Gupta
25